$L^1$-Cauchy sequence of step functions that converges a.e. also converges in $L^1$

Question

Let $X$ be a measure space and $E$ be a Banach space. A step function from $X$ to $E$ is a measurable function ($E$ takes the Borel measure) with finite image and whose support has finite measure.

If $(f_n)$ is a Cauchy sequence of step functions that converge a.e. to a function $f$, then do they also converge to $f$ in the $L^1$ seminorm?

The context for this question is that Lang seems to make this leap in his proof of Theorem 3.4 on page 133 of Real and Functional Analysis, and I want to know if the proof is still correct.

Answer 1

Cauchy in what sense? If they are pointwise Cauchy the here is a counterexample. On the real line with Lebesgue measure let $f_n= \sum\limits_{k=1}^{n} \frac 1 k I_{(k,k+1)}$. Then $f_n \to f=\sum\limits_{k=1}^{\infty} \frac 1 k I_{(k,k+1)}$ but $f_n$ does not tend to $f$ in $L^{1}$ because $\int |f_n-f|=\infty$ for each $n$.

If $(f_n)$ is Cauchy in $L^{1} (X,B)$ then it converges to some $g$ in $L^{1}(X,B)$ by completeness of the latter and $f=g$ in view of the fact that some subsequence of $\|f_n -g\| \to 0$ almost everywhere.

Answer 2

As Kavi Rama Murthy points out, if the sequence is Cauchy in $L^1$, then it converges to $f$ in $L^1$. He has provided a counterexample in the case that the sequence $\{ f_n \}$ is merely pointwise Cauchy. In his example, the pointwise limit $f$ is not in $L^1$. However, it is possible to construct a counterexample in which the limit function is in $L^1$. One such example is given by setting $f_n = \frac{1}{n}I_{[0,n]}$, where $I_{[0,n]}$ is the indicator function for the interval $[0,n]$. The sequence $f_n$ is (uniformly) Cauchy because if $n,m \geq N$, then $|f_n(x) - f_m(x)| \leq \frac{2}{N}$ for all $x$. The sequence tends pointwise to the zero function but does not converge to $0$ in $L^1$ because $\| f_n \|_1 = 1$ for al $n$.